Question:
Balance the equation for the following reaction taking place in aqueous acid solution: Cr_2O^2-_7+ I_2 \rightarrow Cr^3++ IO^-_3
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Solution:
The equation in this problem involves both an oxidation and a reduction reaction. It can be balanced by using the following rules: (1) Separate the net reaction into its two major components, the oxidation process (the loss of electrons) and the reduction process (the gain of electrons). For each of these reactions, balance the charges by adding H^+, if the reaction is occurring in an acidic medium, or OH^- in a basic medium. (2) Balance the oxygens by addition of H_2O. (3) Balance hydrogen atoms by addition of H. (4) Combine the two half reactions, so that all charges from electron transfer cancel out. These rules are applied in the following example. The net reaction is Cr_2O^2-_7+ I_2 \rightarrow Cr^3+ + IO^-_3 The oxidation reaction is I^0_2 + 2lO^-_3+ 10e^- The I atom went from an oxidation number of O in I_2 to + 5 in IO^-_3 , because O always has a - 2 charge. You begin with I_2, therefore, 2 moles of IO^-_3 must be produced and 10 electrons are lost, 5 from each I atom. Recall, the next step is to balance the charges. The right side has a total of 12 negative charges. Add 12 H^+'s to obtain I_2 + 2lO^-_3+ 10e^- +12H^+ To balance the oxygen atoms, add 6H_2O to the left side, since there are 6 O's on the right, thus, I_2 + 6H_2O \rightarrow2lO^-_3 + 10e^- + 12H^+. Hydrogens are already balanced. There are 12 on each side. Proceed to the reduction reaction: Cr_2O^2-_7+ 6e^- \rightarrow 2Cr^3+ Cr began with an oxidation state of + 6 and went to + 3. since 2Cr^3+ are produced, and you began with Cr_2O^2-_7 , a total of 6 electrons are added to the left. Balancing charges: the left side has 8 negative charges and right side has 6 positive charges. If you add 14 H^+ to left, they balance. Both sides now have a net + 3 charge. The equation can now be written. Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3+. To balance oxygen atoms, add 7H_2O `s to right. You obtain Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3++ 7H_2O. The hydrogens are also balanced, 14 on each side. The oxidation reaction becomes I_2 + 6H_2O + 2lO^-_3 + 10e^- + 12H^+ The reduction reaction is Cr_2O^2-_7+ 6e^- + 14H^+ \rightarrow 2Cr^3+ + 7H_2O. Combine these two in such a manner that the number of electrons used in the oxidation reaction is equal to the number used in the reduction. To do this, note that the oxidation reaction has 10e^- and the reduction 6e^-. Both are a multiple of 30. Multiply the oxidation reaction by 3, and the reduction reaction by 5, obtaining oxidation:3I_2 + 18H_2O \rightarrow 6IO^-_3+ 30e^- + 36H^+ reduction:5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O Add these two half-reactions together. 3I_2 + 18H_2O \rightarrow 6IO^-_3+ 30e^- + 36H^+ +5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O 3I_2 + 18H_2O + 5Cr_2O^2-_7 + 30e^- + 70H^+ \rightarrow 10Cr^3+ + 35H_2O + 30e^- + 36H^+ Simplifying, you obtain: 3I_2 + 5Cr_2O^2-_7 + 34H^+ \rightarrow 6IO^-_3 + 10cr^3+ + 17H_2O This is the balanced equation.